A cool video, loved your dedication to the problem even after failing to do so, and seeing you apply new knowledge to attempt to solve it. Do you have discord btw?
My attempt to solve it using the parametric form... I didn't quite finish it... but it feels like the solution is just a few steps away: First dog position is (x(t),y(t)) Second dog position is (1-y(t),x(t)) The speed is 1, so x'^2(t) + y'^2(t) = 1 and given that speed vector is colinear with the second dog - first dog we can say y'/x' = (x-y)/(1-y-x) or, alternatively y' = x' * (x-y)/(1-y-x) we can put it into the first equation to get x' = (1-y-x)/(sqrt(2x(x-1) + 2y(y-1) +1)) and the y' = (x-y)/(sqrt(2x(x-1)+2y(y-1)+1)) this equation isn't very pretty, but we can make it prettier if we shift the origin to (1/2,1/2) x' = - (x+y)/sqrt(2x^2 + 2y^2) y' = (x-y)/sqrt(2x^2 + 2y^2) and even better, to polar coordinate x = rcosθ ; y = rsinθ x' = -cos(θ - pi/4) y' = sin(θ - pi/4) x' = r'cosθ - rsinθ θ' y' = r'sinθ + rcosθ θ' so... we can isolate r' and rθ' r' = -cos(2θ-pi/4) rθ' = sin(2θ-pi/4) and from here... I honesty can't put my finger on it I got a few promising looking expression like r(ln(r) + θ)'/sin(2θ) = - 2sin(3pi/4) or (lnr')^4 - (θ')^4 = cos(4θ - pi/2) but can't integrate it just yet... I feel like... there's a certain transformation that can turn it into something nice
The moment I sent it I realized we can at least find r(θ) let's divide r' by θ' dr/dθ = r cot(2θ - pi/4) dr/r = cot(2θ - pi/4)dθ lnr|r=[r0,r] = ln|sec(2θ-pi/4)| | θ = [-3pi/4;θ] and thus... we have an expression how r depends on the θ in polar coordinate
A cool video, loved your dedication to the problem even after failing to do so, and seeing you apply new knowledge to attempt to solve it. Do you have discord btw?
Great video
good job
My attempt to solve it using the parametric form... I didn't quite finish it... but it feels like the solution is just a few steps away:
First dog position is (x(t),y(t))
Second dog position is (1-y(t),x(t))
The speed is 1, so x'^2(t) + y'^2(t) = 1
and given that speed vector is colinear with the second dog - first dog
we can say
y'/x' = (x-y)/(1-y-x)
or, alternatively
y' = x' * (x-y)/(1-y-x)
we can put it into the first equation to get
x' = (1-y-x)/(sqrt(2x(x-1) + 2y(y-1) +1))
and the
y' = (x-y)/(sqrt(2x(x-1)+2y(y-1)+1))
this equation isn't very pretty, but we can make it prettier if we shift the origin to (1/2,1/2)
x' = - (x+y)/sqrt(2x^2 + 2y^2)
y' = (x-y)/sqrt(2x^2 + 2y^2)
and even better, to polar coordinate
x = rcosθ ; y = rsinθ
x' = -cos(θ - pi/4)
y' = sin(θ - pi/4)
x' = r'cosθ - rsinθ θ'
y' = r'sinθ + rcosθ θ'
so... we can isolate r' and rθ'
r' = -cos(2θ-pi/4)
rθ' = sin(2θ-pi/4)
and from here... I honesty can't put my finger on it
I got a few promising looking expression like
r(ln(r) + θ)'/sin(2θ) = - 2sin(3pi/4)
or
(lnr')^4 - (θ')^4 = cos(4θ - pi/2)
but can't integrate it just yet...
I feel like... there's a certain transformation that can turn it into something nice
The moment I sent it I realized
we can at least find
r(θ)
let's divide r' by θ'
dr/dθ = r cot(2θ - pi/4)
dr/r = cot(2θ - pi/4)dθ
lnr|r=[r0,r] = ln|sec(2θ-pi/4)| | θ = [-3pi/4;θ]
and thus... we have an expression
how r depends on the θ in polar coordinate
👍